A block of mass m slides down an inclined pla | vedclass

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(B) According to the law of conservation of energy,the potential energy lost by the block as it descends a vertical height $h$ is equal to the elastic

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$\sqrt {\frac{{2mgh}}{k}} $

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(B) According to the law of conservation of energy,the potential energy lost by the block as it descends a vertical height $h$ is equal to the elastic potential energy gained by the spring at maximum compression $x$.The loss in gravitational potential energy of the block is $\Delta U_g = mgh$.The gain in elastic potential energy of the spring is $\Delta U_s = \frac{1}{2} Kx^2$.Equating the two:$mgh = \frac{1}{2} Kx^2$Solving for $x$:$x^2 = \frac{2mgh}{K}$$x = \sqrt{\frac{2mgh}{K}}$

$A$ block of mass $m$ slides down an inclined plane as shown in the figure and hits a spring at the bottom,causing it to compress. If the spring constant is $K$,what is the maximum compression of the spring?

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